In the right ∆ABC (m∠C = 90°) the acute angles are in the ratio 5:1, i. e. m∠BAC : m∠ABC = 5:1. If

CH

is the altitude to

AB

and

CL

is the angle bisector of ∠ACB, find m∠HCL.

The ratio is 5:1, and the two angles make a sum of 90, so m∠B=15, and m∠A=75

CL bisects ∠ACB, and ∠ACB=90, so m∠BCL=half of 90=45

m∠CLA, as an exterior angle of ΔBCL, equals m∠B+m∠BCL=15+45=60

CH is the altitude to AB, so m∠CHL=90

m∠HCL=180-m∠CHL-m∠CLH=180-90-60=30

30 degree is the answer.